Suppose that y varies directly as the square root of x, and that y=42 when x=400 . What is y when x=96 ? Round your answer to two decimal places if necessary.

Answer 1

Since y varies directly as the square root of x, you can write the relationship as y = k√x, where k is a constant.

First, you can use the given values (y = 42 and x = 400) to find k:

42 = k√400

42 = k x 20

42/20 = k

k = 2.1

Next, you can use the value of k(2.1) to find y when x = 96:

y = k√x

y = 2.1√96

Now you can solve for y (and round to two decimal places)!

Answer 2

42 / √400 = y / √96

Can you solve for y and answer?

Answer 3

y = ksqrx

42 = ksqr400 = 20k

k = 42/20 = 21/10 = 2.1

y = 2.1sqrx= 2.1sqr96 = 2.1(4)sqr6

= 8.4sqr6

= about 20.57571384

= about 20.58 rounded to 2 decimals

Answer 4

y = k*sqrt(x)

42 = k*sqrt(400)

42 = 20k

42/20 = k = 2.1

y = 2.1*sqrt(96)

96 = 6 * 16, a perfect square, which can be pulled out of the square root as 4.

y = 4 * 2.1 * sqrt(6) = 8.4*sqrt(6) = about 20.58